Question

find the equation of the straight line which passes through the point 4,3 and parallel to line 3x +4y=7​

Answer 1

Answer:

3x + 4y - 24 = 0

Step-by-step explanation:

Let the equation for that line be ax + by + c = 0

Given,

ax + by + c is parallel to 3x + 4y = 7.

It means :

= > Slope of ax + by + c = 0 is equal to the slope of 3x + 4y = 7

= > Slope of ax + by + c = 0 is given by m from the expression given here : 3x + 4y = 7 = > 4y = 7 - 3x = > y = ( - 3 / 4 )x + ( 7 / 3 ) = > On comparing with y = mx + c, m = slope of '3x + 4y 7' = - 3 / 4.

= > Slope of ax + by + c = 0 is - 3 / 4.

Therefore,

= > Equation of that line

= > y - 3 = ( - 3 / 4 )( x - 4 )

= > 4y - 12 = - 3x + 12

= > 3x + 4y - 24 = 0.

Answer 2

Question :-- find the equation of the straight line which passes through the point 4,3 and parallel to line 3x +4y=7 ?.

Formula and Concept used :--

→ Equations of straight lines are in the form y = mx + c where, we have m is the slope of line .

→ when lines are parallel their slope is Equal.

→ Equation of line with point (x1,y1) is given by (y-y1) = m(x-x1) ..

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Solution :--

First lets Find the Slope of line 3x+4y = 7

→ 3x + 4y = 7

→ 4y = (-3x) + 7

Dividing both sides by 4 ,

→ y = (-3/4)x + 7

Comparing it with y = mx + c, we get,

→ m = (-3/4) = Slope of line .

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Now, it is said that, The straight line is parallel to This line.

That means, Slope of Required line will also be (-3/4).

And, Required line will pass Through (4,3) .

So, we have,

→ x1 = 4 , and y1 = 3, m = (-3/4)

Equation of line = (y-y1) = m(x-x1)

putting values we get,

→ (y - 3) = (-3/4) [ x - 4 ]

→ 4(y - 3) = (-3)(x-4)

→ 4y - 12 = (-3)x + 12

→ 4y + 3x = 12 + 12

→ 3x + 4y = 24

→ 3x + 4y - 24 = 0.

Hence, The Equation of Straight line which passes through the given point and parallel to 3x+4y = 7 is 3x + 4y = 24 .

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